In Young's double slit experiment, the ratio of maximum and minimum intensities in the fringe system is $$9: 1$$. The ratio of amplitudes of coherent sources is

In the young's double slit experiment the fringe width of the interference pattern is found to be $$3.2 \times 10^{-4} \mathrm{~m}$$, when the light of wave length $$6400^{\circ} \mathrm{A}$$ is used. What will be change in fringe width if the light is replaced with a light of wave length $$4800^{\circ} \mathrm{A}$$

A light having wavelength $$6400^{\circ} \mathrm{A}$$ is incident normally on a slit of width $$2 \mathrm{~mm}$$. Then the linear width of the central maximum on the screen kept $$2 \mathrm{~m}$$ from the slit is :

In Young's double slit experiment, the two slits are separated by 0.2 mm and they are 1 m from the screen. The wavelength of the light used is 500 nm. The distance between 6th maxima and 10th minima on the screen is closest to